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2 years ago

What is a neutralization reaction?

Chemistry

2 answers:

Aleks [24]2 years ago

6 0

Answer:

D or A your choice

Explanation:

uhhh didnt u- nvm

A neutralization reaction is when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate wate

vazorg [7]2 years ago

3 0

The answer is D. A reaction that removes essentially all H+ and OH-

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2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

2 years ago

Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo

Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

4 0

3 years ago

[Ar]4s23d104p3 is the electron configuration of a(n) ________ atom.

madreJ [45]

<span>[Ar]4s23d104p3 is the electron configuration of a(n) __As______ atom.</span>

3 0

3 years ago

Read 2 more answers

In which of the following substances does sharing of electrons between two atoms occur?

malfutka [58]

Answer:- B- $SO_4^-^2$

Explanations:- Sharing of electrons takes place between non metals and the bond formed by the sharing of electrons is known as covalent bond. In first choice we only have a zinc metal and so no sharing of electrons would be possible here.

In third choice we have NaCl. Na is a metal and Cl is a non metal. The property of metals is to transfer its valence electrons to the non metal and the bond formed is known as ionic bond. So, third choice is also not correct.

In choice B we have sulfate ion that has sulfur and oxygen atoms and both of these are non metals. So, sharing of electrons is present here between the S and O atoms and covalent bonds are formed.

So, choice B- $SO_4^-^2$ is the right answer.

6 0

2 years ago

In the laboratory you dilute 2.49 ml of a concentrated 6.00 m hydrochloric acid solution to a total volume of 50.0 ml. what is t

Marizza181 [45]

Equation: M1V1 = M2V2

Where M = concentration & V = volume

Step 1: Write down what is given and what you are trying to find

Given: M1 = 6.00M, V1 = 2.49mL, and V2 = 50.0mL
Find: M2

Step 2: Plug in the values into the equation

M1V1 = M2V2
(6.00M)(2.49mL) = (M2)(50.0mL)

Step 3: Isolate the variable (Divide both sides by 50.0mL so M2 is by itself)

(6.00M)(2.49mL) / (50.0mL) = M2

Answer: M2 = 0.30M
*Don't forget sig figs & units!

4 0

2 years ago